Tweaker's Asylum

• Re: excessive highs customizer II addendum madalo 22:46:08 07/24/00 (0)
  In Reply to: Re: excessive highs customizer posted by madalo on July 21, 2000 at 13:06:18:

  Ron,

  When I re-read what I posted I reallized I went through it kinda fast, and may be some things unclear. Such as jumping from calculating a capacitor value to changing a resistor value...

  Im sorry if that transition seems incomplete. The example I had was only for the 8ohm resistor and how to calculate its capacitor for the original example cross-over frequency of 4000Hz.

  Now when you are changing resistor values, that's another story all together. Remember if your capacitance remains constant, & your resistor changes, it will also change your cross-over frequency. This is one factor that you must consider when you make your listening tests. If tuning down the cross-over frequency slightly supplements the tweeter attenuation, then by all means don't replace the capacitor. On the other hand, you might want to keep a small selection of caps on hand to do little comparisons of various cross-over points, and/or the original cross-over point with various resistance values.

  An important property of an L-pad, is that they have the ability to present a series resistance to the tweeter, but also an 8ohm resistive load to the capacitor, if wired in as their manufacturers suggest, instead of just a series resistor/rheostat. This advantage is that your tweeter gets its series attenuation resistor, while the capacitor has a much better time maintaining the original cross-over frequency.

  There's enough information there, I think, to construct a formula, to determine a resistor at a particular attenuation level, at a specified frequency, but my algebra skills aren't as up to snuff as they used to be. How are you at logorithms? Any takers? Meanwhile I'm still looking for the book with that formula.

  Take care.

  Dan
  
  

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• excessive highs customizer II madalo 16:40:12 07/24/00 (0)
  In Reply to: Re: excessive highs customizer posted by madalo on July 21, 2000 at 13:06:18:

  Ron,

  I will try to give you an idea to hopefully "fix" your problem. I misplaced my book that had the original formula in it, but I looked up the circuit in another book, and it isn't really presented in the same way. It gives another interesting feature of this circuit - equalization! But for now, this is basically what happens:

  Let me assume a few things for clarification of this little explanation (these are arbitrary values)...
  tweeter = 8ohm
  capacitor = 5uF
  series resistor = 8ohm
  cross-over frequency = 4000Hz

  Your tweeter is a reactive load, which is something like a resistor, except that its resistance can change with frequency. We call this type of resistance  "reactance" or "impedance."  At the frequency of your cross-over, the capacitor's (which is also a reactive element) impedance changes to the same value as the impedance of your tweeter; hence putting half of your amplifiers applied power into the tweeter & half of its voltage across the capacitor.

  If you increase the resistance (impedance) that the capacitor sees by adding a series resistor, it will do two things: first - it will lower the cross-over frequency; second - it will place less voltage (& power) across the tweeter. The reason for this is that at the same original frequency, the impedance + resistance of the tweeter/resistor circuit is greater than that of the capacitor (about twice as high) so it still produces the maximum voltage(& power) output across the tweeter. So the frequency has to go lower for the capacitor's impedance to equal that of the tweeter/resistor circuit (which would be at the theoretical new cross-over frequency).

  When I say "maximum" here, it's already different (lower) than your original tweeter output voltage (& power), because the original output voltage (and power) are being shared by both tweeter & resistor. Remember, this is without changing ANY components, only adding a series resistor. It's still the same capacitor - its reactance is still the same.

  Imagine you have a battery, and to the battery you connect a light bulb. Then you connect another identical light bulb in series. Both lights will be on, but the first light bulb will only have half the battery's power and not near as bright! Now imagine you replace the second light bulb with a resistor electrically identical to the second light bulb. You only have one light, but its still much dimmer than the original circuit. I hope that's a clear analogy, it just takes away the more confusing parts of reactances & changing frequencies.

  To add a resistor, and maintain the original cross-over frequency, you must change the value of the capacitor. Since the impedance after the capacitor only exists as the tweeter's impedance + the resistor's resistance we can figure that to be about 16ohm - in this case with my above assumptions. The formula for calculating a capacitor for a high-pass filter is this: 

  C = 1/  2.pi.f.Xc   
  (capacitance = 1 / 2 x pi x frequency x reactance@frequency)

  Xc = about the same as our total impedance - in this case we figured 16 ohm.

  f = the original cross-over frequency

  after calculation

  C = 2.5uF       
  (Notice how the components are mathematically proportional)

  Got the idea? OK, this calculation should theoretically give a -3dB cut (half power) & if you want a -6dB cut you'll have to use 32ohm. My guess is that you wont need to go over 8ohm. I mentioned this before, but I will mention again... you should check the specs of your tweeter, mid, & bass drivers to obtain the sensitivity rating. This might give you an idea as to what resistor values you might be looking for. Remember proportionality is the key when dealing with changing the values.

  You might try even getting two cheap ($12) 8ohm L-pads & wire them in series to each other & your tweeter. Connect them outside your speaker enclosure & you can then adjust your series resistance from 0ohm to 16ohm by dialing the L-pads. This way, you can select the series resistance by what sounds best, instead of relying simply on calculations. Most likely one L-pad will suffice, as I mentioned earlier, you probably won't need to go over 8ohm. After dialing in the right sound, measure the resistance of the L-pad(s) and buy an equivalent resistor to permanently wire in series with your tweeter. Make sure you make your listening tests before and after changes, and with the same recorded material & volume levels.

  I mentioned sticking two L-pads in series , this could be substituted as one L-pad for 0ohm to 8ohm, then use one L-pad + one 8ohm resistor to listen to 8ohm through 16ohm. The advantage of having two in series is that you don't need to rewire anything to make continuous variable listening tests. It's disadvantage is that you can't test that way in stereo. You may be best off getting two L-pads (at first) and testing one speaker with them in vs. out of the tweeter circuit. then if you find it sounds best below 8 ohm (which I'm sure will be the case), then wire the second L-pad in series with the other tweeter & re-test in stereo vs. with them out of the circuit in stereo.

  Ideally, you would want to find a resistor which can handle the same amount of power as your tweeter, but such a large resistor may not be practical. You can generally drive a resistor quite hard without damage, so around 10W should suffice unless you listen to your music extremely loud for extremely long periods of time. There are some resistors which use an aluminum body that can handle the same amounts power in a smaller package. You can buy some that may be previously used much cheaper, & if you buy several of the same value, you can measure them to find the best matching pair which come closest to your calculated value.

  Now onto the other benefit of this circuit - Equalization... It so happens in the book that I actually DID manage to find, it pitches the circuit as being able to "boost upper-highs" but I haven't found any electrical theory to explain this beyond what is only seemingly a "boost." To go into further detail: As I've mentioned, a tweeter is a reactive element, and in upper-high frequencies the coil inductance begins to play a part in rolling off highs. The characteristics of adding a series resistor to the tweeter circuit causes the reactance of the coil to have a much lower effect on the tweeter circuit as a whole. Therefore, you will get a much more even impedance as seen by your capacitor & a much more even frequency response. That doesn't mean that you won't receive the over-all cut that I explained earlier, it just means that the droopy upper-highs (if that's a characteristic of your tweeter) will simply be ironed-out in respect to the rest of the output response of your tweeter. In other words, it evens out the top-end & bottom-end of your tweeter's response, partially by increasing the linearity of the impedance curve & partially by lowering the over all output of the tweeter. The other selling-point which they mentioned in my book was it "can balance the tweeter to the woofer." This matching is basically the cut which we calculated earlier. Remember when I was mentioning checking the sensitivity specification of  your drivers? Well suppose your woofer is 89dB & your tweeter is 92dB... a -3dB cut would put them right on par with each other! Built-in equalization! across the board.

  Let me know how it Goes!

  -Dan
  
  

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